laplacetransform 2t^2-t+1

解答

拉普拉斯变换

2 t^{2} - t + 1

\frac{4}{s ^{3}} - \frac{1}{s ^{2}} + \frac{1}{s}

求解步骤

L {2 t^{2} - t + 1}

利用拉普拉斯变换的线性特性:

对于函数

f (t), g (t)

和常数

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= 2 L {t^{2}} - L {t} + L {1}

L {t^{2}} : \frac{2}{s ^{3}}

L {t} : \frac{1}{s ^{2}}

L {1} : \frac{1}{s}

= 2 \cdot \frac{2}{s ^{3}} - \frac{1}{s ^{2}} + \frac{1}{s}

整理

2 \frac{2}{s ^{3}} - \frac{1}{s ^{2}} + \frac{1}{s} : \frac{4}{s ^{3}} - \frac{1}{s ^{2}} + \frac{1}{s}

= \frac{4}{s ^{3}} - \frac{1}{s ^{2}} + \frac{1}{s}

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